# Interpreting the Einstein stress-energy tensor T_ab

1. Jan 23, 2007

### cesiumfrog

The Einstein field equation relates the curvature of space to the distribution of matter, representing the latter with a tensor T. The components of this tensor have been interpreted as representing the volume-density of mass-energy together with the "pressure" in each spatial direction.

Can someone explain this pressure? How does one calculate the component $T_{22}$ for a classical distribution of (ideal) gas or fluid? I assume the density component is trivially just the classical mass density, are the pressure components equal to the outward force per unit area?

2. Jan 23, 2007

### Crosson

If the ideal gas is uniform with pressure $$P$$ then we have $$T_{22}=T_{33}=T_{44} = P$$.

3. Jan 23, 2007

### pervect

Staff Emeritus
It may not be obvious at first glance, but it takes a rank 2 tensor to describe the force / unit area in general. Pressure is not a scalar, nor a vector. This is just as true in engineering (which also uses a version of the stress energy tensor) as it is in physics.

When you have a standard orthonormal coodinate system (i.e. x,y,z) then the components of the stress-energy tensor are just the pressures. So the stress-energy tensor in such a coordinate system consists of energy and momentum terms "tacked onto" a classical 3x3 pressure (or stress) tensor.

If the coordinates are globally not orthonormal, one can still construct a local orthonormal frame field. Imagine, for instance, polar coordinates. Globally, you have r and $\theta$. Locally, you have unit vectors $\hat{r}$ and $\hat{\theta}}$. Most physics textbooks will use the notation $$T_{\hat{r}{\hat{\theta}}$$ to describe the stress-energy tensor in such a frame field. This is closest to the engineering usage as nearly as I can determine (but I'm not extremely familiar with mechanical engineeering).

There is another difference between the physicists usgae and the engineering usage. If you have a rotating wheel, the engineering usage is to compute the stress-energy tensor by default in a frame-field moving along with the rotating disk, while the physicists usage is to consider the stationary frame field by defualt.

These differences can be important if you are unaware of them, but fundamentally the mathematics is the same.

You can think of positive terms in the stress-energy tensor as pressures - and negative terms as tension. If you have a static rope supporting a weight, for instance, with the rope running up and down in the z direction, T_zz would be negative, and T_xx and T_yy and the diagonal terms would all be zero.

4. Jan 23, 2007

### cesiumfrog

I understand however that this tensor can, always, be diagonalised. Does this mean pressure is a vector in the rest frame of the fluid?

So is dark energy exactly equivalent to a meta-material consisting of a web of interconnected and stretched springs?

5. Jan 24, 2007

### pervect

Staff Emeritus
Not exactly. Think of the moment of inertia tensor. It can also be diagonalized. Physically, the moment of inertia tensor can be represented by the ellipsoid of inertia. (Have you heard of that, too? It saves some explanations if you have).

The same thing is true of pressure, you can think of a "pressure ellipsoid" in cases where you have anisotropic pressures. When you have anisotropic pressures, you will have "principal axes" of the pressure. The stress-energy tensor gives the orientation of these axes in the particular coordinate system of choice, as well as the magnitude of the pressures (or tensions) along these three principal axes. The orientation is given by the specific rotation you use to diagonalize the matrix, and the magnitude of the pressures are given by the values of the diagonal elements.

I suppose that's one way of looking at it. Of course, springs don't violate the weak/strong/etc energy conditions the way dark energy does.

6. Jan 24, 2007

### quantum123

Good work, pervect.
I always appreciate your science within the maths, maths supporting the science approach.
Now you even bring in engineering.
I like it! :)
I have actually seen a book with heavy tensor analysis, stress-momentum-energy etc. Guess what: it is a biomedical-engineering text book, they want to know what is inside a blood vesssl, in a living cell, in an organ etc.

7. Jan 24, 2007

### cesiumfrog

That seems inconsistant: if this meta-material is equivalent to dark energy then surely both must satisfy/violate the same conditions? I understand that dark energy (at least in the form of positive cosmological constant) does satisfy most (ie. weak, dominant and null) energy conditions.. what does it mean physically for something to violate the strong energy condition?

8. Jan 24, 2007

### pervect

Staff Emeritus
Bah. I was in too much of a hurry when I tried to clarify my previous post, which was also not very good with regards to the energy conditions.

Basically what I was trying to say is that dark matter is one form of exotic matter. However, my now-deleted example of what this means wasn't correct.

Last edited: Jan 25, 2007
9. Jan 25, 2007

### cesiumfrog

... huh?

10. Jan 25, 2007

### pervect

Staff Emeritus
I've got some more time, so I'm going to redo what I think is an interesting analysis, but this time around get the right answer :-)

Suppose we have an infinite region of "true vacuum" with a stress energy tensor of zero, and inside this infinite region of true vacuum, we have a small spherical "box" of some normal material, and we fill this box with dark energy with rho (the energy density) = -P (the isotropic pressure) = constant.

(We need the box - we can't have a stable static boundary between a true vacuum and the dark matter without it).

Because this is a static geometry in a true vacuum, so we can apply the concept of Komar mass, which is basically a volume intergal of rho+Pxx+Pyy+Pzz.

See http://en.wikipedia.org/wiki/Mass_in_general_relativity[/url] and [url]http://en.wikipedia.org/wiki/Komar_mass [Broken].

(Disclaimer: I'm the primary author of both of these articles, so they shouldn't be considered a totally independent source, though I belive Chris Hillman and hopefully some Wikipedians have looked them over for errors.)

If we use the Komar formula to find the mass of the "empty" box, and then use the Komar formula to find the mass of the box full of dark matter, we will find that the mass increases when we add the dark matter.

The reason for this is that the pressure terms will cancel out. Without the dark matter, there is no pressure in the walls of the box. When we add the dark matter, there is a negative pressure inside the box. There is still zero pressure outside the box. The net result is a compression of the box when we add the dark matter to the interior. It can be shown that the integral of this positive pressure in the walls of the box caused by this compression is equal and opposite to the integral of the negative pressure over the volume of the interior of the box. I won't go through the details here, see for instance

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm

and perform the appropriate integration.

Since the pressure terms cancel out, this leaves only the energy term, and because dark matter has a positive energy density, the mass of the box will increase when we add dark matter to it.

However, we examine the "gravitational field" (by which I mean the proper acceleration of a static observer) just inside the box, the walls of the box will not contribute, and the pressure terms show their importance.

This if we place a small test particle just inside the walls of the box, it will move away from the center of the box unless held in place by an inward pointing force. We have effectively a gravitationally repulsive force due to the dark matter.

This is because the Komar integral for the mass of just the dark matter is negative. It's negative because the pressure is isotropic, and rho+3P is negative (rho+P =0, so rho+3P < 0).

Last edited by a moderator: Apr 22, 2017 at 3:22 PM
11. Jan 27, 2007

### cesiumfrog

Obviously I've spent too much time only thinking about vacuum solutions.. just to clarify, is there a particular reason you use the terms "dark energy" and "dark matter" interchangeably?

12. Jan 27, 2007

### MeJennifer

I thought dark matter was supposed to explain the gravitational peculiarities of galaxies and dark energy would explain the accelerating expansion.

At one point in the future we have to start thinking of how far we can stretch the validity of GR. Sure we can just postulate that GR is eternally right and every future anomality is, with a patronizing voice and a straight face of course, "explained" by yet another dark thing.

13. Jan 27, 2007

### quantum123

That is precisely the point the Internet community is sick of.
Recently some bright halo was discovered around the andromeda galaxy, some people jokingly say could again be attributed dark matter. Who knows what bright matter that present telescopes are not strong enough to see, yet attributed to dark matter, and then later discovered because we have the patience and more powerful telescopes.

14. Jan 27, 2007

### pervect

Staff Emeritus
Did I do that? I hope not. But the way things have been going this week, I wouldn't be surprised if I did :-(. I'm afraid RL has been interfering with my PF posting.

Anyway, there are actual several forms of dark energy, the one I was talking about was the sort associated with a cosmological constant. This is the sort that has a postive energy density rho, and a negative pressure, with P = -rho in geometric units. There are some alternate forms with different "equations of state" (i.e. pressure vs density relation) which share the general features of a positive energy density and a gravitationally significant negative pressure.

Dark energy is different from dark matter. There are several possible forms of dark matter, dark matter could be any form of matter that doesn't emit visible radiation, including for example MACHO's and wimp's.

Dark matter isn't exotic matter. While dark matter has not been directly observed, I belive that as the term as it is usually used implies that one is talking about "cold dark matter" which implies that the pressure terms are negligible. Thus one would assume that rho=positive and P $\approx$ 0 for dark matter - i.e. that the pressure is zero, or at least gravitationally insignificant.

This is different from dark energy, which is assumed to fuel the apparently accelerated expansion of the universe and has a gravitationally significant negative pressure.

Last edited: Jan 27, 2007
15. Jan 30, 2007

### cesiumfrog

OK. So, satisfying the strong energy condition means that if several very distant observers each use a string to a dangle giant coconut-shell over the non-vacuum region, then by measuring the pull they will all make consistant deductions of the mass.

I find this strong condition quite unmotivating so far, partly because the "observers at infinity" bring my mind to the problems defining of an event horizon (opposed perhaps to some form of local horizon) in Schwarzschild versus flat/Rindler spacetime, and partly because current experimental evidence seems to indicate our universe possesses dark energy that violates this condition. On the other hand, it seems the other energy conditions can be equated to protecting local causality, and are not quite violated by the standard dark energy.

But back to understanding the physical interpretation of the tensor (and the gap in equivalence to my spring-web), can (or has?) it actually been shown (proven?) that "normal" (baryonic?) matter satisfies the various conditions and that any distribution thereof also still does?

16. Feb 7, 2007

### cesiumfrog

From another thread (quoting here to keep both on-topic):
Isn't the net momentum flow usually zero?

For example, a gas may be confined as to have high pressure, but on average (and in every spacelike direction) there is a particle flowing into any volume for every particle that flows out.

If it is actually the momentum flux on which $T_{ab}$ depends, shouldn't pressure contribute nothing (unless the fluid is let expand)? Or rather (since the pressure should nonetheless increase the potential/mass-energy of the fluid), shouldn't it affect only the $T_{00}$ component (in coordinates of the fluid's rest frame) and contribute nothing to (say) $T_{22}$? Or is this incorrect?

17. Feb 9, 2007

### pervect

Staff Emeritus
Baez, for one, does call pressure the rate at which x-momentum is transported in the x direction (in his GR tutorial, I can give the link if you like), but for the reasons you cite I don't think that definition of pressure is particularly clear.

If one uses a "swarm of particle" models, when all particles at a point move in the same direction, one can say that the pressure at that point is zero.

If one considers a swarm of particles where half the particles are moving to the right, and half the particles are moving to the left, there is an anisotropic pressure in the left-right direction.

In an ideal gas or ideal fluid, particles all move isotropically in all directions, and the pressure is a constant independent of direction.The higher the average speed of the particle, the higher the pressure.

In an ideal gas or ideal fluid, the stress-energy tensor is diag(rho,P,P,P)
where rho is the energy density per unit volume, and P is the pressure.

18. Feb 9, 2007

### pervect

Staff Emeritus
I should mention another familiar approach to defining pressure which is very simple. You build a small cubical box, and you measure the normal force / unit area (i.e. the force pointing inwards or outwards) on all sides of the box, due to the contents of the box (matter, energy, etc). If you align the box so that the principal axes of the box are the principal axes of the pressure tensor, you've completely characterized the pressure of the box by the pressure on the three walls (and by the alignment of the box).

If you don't align the box, the analysis gets more complicated, but you can with enough effort make the force / unit area concept work. The force/unit area becomes equal to a general second rank tensor multiplied by a normal vector. Only when you align the box properly does the general second rank tensor become diagonal.

See for instance http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/Sec12.html [Broken]